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Miscellaneous

Permutations and Combinations

**Example-1 :-** How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE ?

In the word INVOLUTE, Vowels = {I, O, E, U} = 4 and Consonants = {N, V, L, T} = 4. The number of ways of selecting 3 vowels out of 4 = ⁴C₃ = 4. The number of ways of selecting 2 consonants out of 4 = ⁴C₂ = 6. Therefore, the number of combinations of 3 vowels and 2 consonants is 4 × 6 = 24. Now, each of these 24 combinations has 5 letters which can be arranged among themselves in 5 ! ways. Therefore, the required number of different words is 24 × 5 ! = 2880.

**Example-2 :-** A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has

(i) no girl ?

(ii) at least one boy and one girl ?

(iii) at least 3 girls ?

(i) Since, the team will not include any girl, therefore, only boys are to be selected. 5 boys out of 7 boys can be selected in ⁷C₅ ways. Therefore, the required number of ways = ⁷C₅ = 7!/5!2! = (6 x 7)/2 = 21

(ii) Since, at least one boy and one girl are to be there in every team. Therefore, the team can consist of 1 boy and 4 girls can be selected in ⁷C₁ × ⁴C₄ ways. 2 boys and 3 girls can be selected in ⁷C₂ × ⁴C₃ ways. 3 boys and 2 girls can be selected in ⁷C₃ × ⁴C₂ ways. 4 boys and 1 girl can be selected in ⁷C₄ × ⁴C₁ ways. Therefore, the required number of ways = ⁷C₁ × ⁴C4 + ⁷C₂ × ⁴C₃ + ⁷C₃ × ⁴C₂ + ⁷C₄ × ⁴C₁ = 7 + 84 + 210 + 140 = 441

(iii) Since, the team has to consist of at least 3 girls, the team can consist of following below : Note that the team cannot have all 5 girls, because, the group has only 4 girls. 3 girls and 2 boys can be selected in ⁴C₃ × ⁷C₂ ways. 4 girls and 1 boy can be selected in ⁴C₄ × ⁷C₁ ways. Therefore, the required number of ways = ⁴C₃ × ⁷C₂ + ⁴C₄ × ⁷C₁ = 84 + 7 = 91

**Example-3 :-** Find the number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the 50th word?

There are 5 letters in the word AGAIN, in which A appears 2 times. Therefore, the required number of words = 5!/2! = 60 To get the number of words starting with A, we fix the letter A at the extreme left position, we then rearrange the remaining 4 letters taken all at a time. There will be as many arrangements of these 4 letters taken 4 at a time as there are permutations of 4 different things taken 4 at a time. Hence, the number of words starting with A = 4! = 24. Then, starting with G, the number of words 4!/2! = 12 as after placing G at the extreme left position, we are left with the letters A, A, I and N. Similarly, there are 12 words starting with the next letter I. Total number of words so far obtained = 24 + 12 + 12 = 48. The 49th word is NAAGI. The 50th word is NAAIG.

**Example-4 :-** How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4?

The number of 7-digit arrangements, clearly, 7!/3!2! = 420. But, this will include those numbers also, which have 0 at the extreme left position. The number of such arrangements 6!/3!2! (by fixing 0 at the extreme left position) = 60. Therefore, the required number of numbers = 420 – 60 = 360.

**Example-5 :-** In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?

Let us first seat the 5 girls. This can be done in 5! ways. For each such arrangement, the three boys can be seated only at the cross marked places. × G × G × G × G × G ×. There are 6 cross marked places and the three boys can be seated in 6P3 ways. Hence, by multiplication principle, the total number of ways = 5! × ⁶P₃ = 5! × 6!/3! = 5! x (6 x 5 x 4 x 3!)/3! = 5 x 4 x 3 x 2 x 1 x 6 x 5 x 4 = 14400.

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